Question

What is the greatest number of 3 digit which when divided by 6,9 and 12 leaves a remainder 3 in each case?

Answer 1

step 1- find L.C.M of given number

L.C.M of 6,9 and 12 is 36
step 2- find largest 3 digit number is divisible by 36
largest 3 digit number is 999 and largest 3 digit number is divisible by 36 is 972
step3-add the remainder to the number
972+3=975

Answer 2

Answer:

Required Number is 975.

Step-by-step explanation:

Given: Required Number when divided by 3 , 9 and 12 leaves 3 as remainder in each case.

To find: Largest 3-Digit Number.

First we find LCM of 6 , 9 & 12 then we find multiple of LCM to obtain largest 3-Digit Number.

LCM of 6 , 9 , and 12

6 = 2 × 3

9 = 3 × 3

12 = 2 × 2 × 3

LCM = 3 × 2 × 2 × 3 = 36

Now, we obtain largest 3-Digit number

Multiple of 36 = 36, 72, 108, 144, 180, 216, 252, 288, 324, 360, 396, 432, 468, 504, 540, 576, 612, 648, 684, 720, 756, 792, 828, 864, 900, 936, 972, 1008, ...

⇒ Largest 3-digit number divisible by 6 , 9 , & 12 = 972

Since we get 3 as remainder. So, required number = 972 + 3 = 975

Therefore, Required Number is 975.